HNO3 + NH3 = NH4NO3

HNO_3 (nitric acid) + NH_3 (ammonia) ⟶ NH_4NO_3 (ammonium nitrate)

Balance the chemical equation algebraically: HNO_3 + NH_3 ⟶ NH_4NO_3 Add stoichiometric coefficients, c_i, to the reactants and products: c_1 HNO_3 + c_2 NH_3 ⟶ c_3 NH_4NO_3 Set the number of atoms in the reactants equal to the number of atoms in the products for H, N and O: H: | c_1 + 3 c_2 = 4 c_3 N: | c_1 + c_2 = 2 c_3 O: | 3 c_1 = 3 c_3 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | HNO_3 + NH_3 ⟶ NH_4NO_3

+ ⟶

nitric acid + ammonia ⟶ ammonium nitrate

| nitric acid | ammonia | ammonium nitrate molecular free energy | -80.7 kJ/mol | -16.4 kJ/mol | -183.9 kJ/mol total free energy | -80.7 kJ/mol | -16.4 kJ/mol | -183.9 kJ/mol | G_initial = -97.1 kJ/mol | | G_final = -183.9 kJ/mol ΔG_rxn^0 | -183.9 kJ/mol - -97.1 kJ/mol = -86.8 kJ/mol (exergonic) | |

Construct the equilibrium constant, K, expression for: HNO_3 + NH_3 ⟶ NH_4NO_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the activity expression for each chemical species. • Use the activity expressions to build the equilibrium constant expression. Write the balanced chemical equation: HNO_3 + NH_3 ⟶ NH_4NO_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HNO_3 | 1 | -1 NH_3 | 1 | -1 NH_4NO_3 | 1 | 1 Assemble the activity expressions accounting for the state of matter and ν_i: chemical species | c_i | ν_i | activity expression HNO_3 | 1 | -1 | ([HNO3])^(-1) NH_3 | 1 | -1 | ([NH3])^(-1) NH_4NO_3 | 1 | 1 | [NH4NO3] The equilibrium constant symbol in the concentration basis is: K_c Mulitply the activity expressions to arrive at the K_c expression: Answer: | | K_c = ([HNO3])^(-1) ([NH3])^(-1) [NH4NO3] = ([NH4NO3])/([HNO3] [NH3])

Construct the rate of reaction expression for: HNO_3 + NH_3 ⟶ NH_4NO_3 Plan: • Balance the chemical equation. • Determine the stoichiometric numbers. • Assemble the rate term for each chemical species. • Write the rate of reaction expression. Write the balanced chemical equation: HNO_3 + NH_3 ⟶ NH_4NO_3 Assign stoichiometric numbers, ν_i, using the stoichiometric coefficients, c_i, from the balanced chemical equation in the following manner: ν_i = -c_i for reactants and ν_i = c_i for products: chemical species | c_i | ν_i HNO_3 | 1 | -1 NH_3 | 1 | -1 NH_4NO_3 | 1 | 1 The rate term for each chemical species, B_i, is 1/ν_i(Δ[B_i])/(Δt) where [B_i] is the amount concentration and t is time: chemical species | c_i | ν_i | rate term HNO_3 | 1 | -1 | -(Δ[HNO3])/(Δt) NH_3 | 1 | -1 | -(Δ[NH3])/(Δt) NH_4NO_3 | 1 | 1 | (Δ[NH4NO3])/(Δt) (for infinitesimal rate of change, replace Δ with d) Set the rate terms equal to each other to arrive at the rate expression: Answer: | | rate = -(Δ[HNO3])/(Δt) = -(Δ[NH3])/(Δt) = (Δ[NH4NO3])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| nitric acid | ammonia | ammonium nitrate formula | HNO_3 | NH_3 | NH_4NO_3 Hill formula | HNO_3 | H_3N | H_4N_2O_3 name | nitric acid | ammonia | ammonium nitrate

| nitric acid | ammonia | ammonium nitrate molar mass | 63.012 g/mol | 17.031 g/mol | 80.04 g/mol phase | liquid (at STP) | gas (at STP) | solid (at STP) melting point | -41.6 °C | -77.73 °C | 169 °C boiling point | 83 °C | -33.33 °C | 210 °C density | 1.5129 g/cm^3 | 6.96×10^-4 g/cm^3 (at 25 °C) | 1.73 g/cm^3 solubility in water | miscible | | surface tension | | 0.0234 N/m | dynamic viscosity | 7.6×10^-4 Pa s (at 25 °C) | 1.009×10^-5 Pa s (at 25 °C) | odor | | | odorless